DISINI 2014CSMCSolution
The CENTRE for EDUCATION
in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
2014 Canadian Senior
Mathematics Contest
Thursday, November 20, 2014
(in North America and South America)
Friday, November 21, 2014
(outside of North America and South America)
Solutions
©2014 University of Waterloo
2014 Canadian Senior Mathematics Contest Solutions
Page 2
Part A
1. We note that ∠DAF = ∠F AB − ∠DAB and that ∠F AB = ∠F AE + ∠EAB.
Since ABCD is a square, then ∠DAB = 90◦ .
Since each of △AEF and △ABE is equilateral, then ∠F AE = ∠EAB = 60◦ .
Therefore, ∠DAF = (60◦ + 60◦ ) − 90◦ = 30◦ .
Answer: 30◦
2. Since the ratio of the number of dimes to the number of quarters is 3 : 2, then we let the
numbers of dimes and quarters be 3k and 2k, respectively, for some positive integer k.
Since each dime is worth 10 cents, then the 3k dimes are worth 10(3k) = 30k cents.
Since each quarter is worth 25 cents, then the 2k quarters are worth 25(2k) = 50k cents.
Since the total value of the coins is $4 which equals 400 cents, then 30k + 50k = 400 or
80k = 400, and so k = 5.
Therefore, there are 3(5) = 15 dimes in the jar.
(Note that 15 dimes are worth 150 cents and 2(5) = 10 quarters are worth 250 cents, for a total
of 400 cents, as required.)
Answer: 15
3. We note that 5000 = 5 · 1000 = 5 · 103 = 5 · (2 · 5)3 = 5 · 23 · 53 = 23 54 .
Since neither m nor n is divisible by 10, then neither m nor n can have factors of both 2 and 5.
Since the product mn equals 5000, the only prime factors of 5000 are 2 and 5, and neither m
nor n has factors of both 2 and 5, then m and n must be 23 and 54 , or 54 and 23 .
Therefore, m + n = 23 + 54 = 8 + 625 = 633.
Answer: 633
4. Solution 1
Since f (x) + f (x + 3) = 2x + 5 for all x, then using x = 2 we obtain f (2) + f (5) = 2(2) + 5
and so f (2) + f (5) = 9.
Also, using x = 5 we obtain f (5) + f (8) = 2(5) + 5, and so f (5) + f (8) = 15.
Adding these two equations, we obtain f (2) + f (8) + 2f (5) = 24.
Since f (8) + f (2) = 12, then 2f (5) = 24 − (f (2) + f (8)) = 24 − 12 = 12 and so f (5) = 6.
Solution 2
The function f (x) = x + 1 satisfies the required properties, since
• f (x + 3) = (x + 3) + 1 = x + 4 and so f (x) + f (x + 3) = (x + 1) + (x + 4) = 2x + 5, and
• f (2) = 2 + 1 = 3 and f (8) = 8 + 1 = 9 gives f (8) + f (2) = 12.
So f (x) = x + 1 is a function that works and the question implies that the value of f (5) is the
same no matter which function f that satisfies the given properties is chosen.
Therefore, f (5) = 5 + 1 = 6.
(Note that there may be other functions that satisfy these relationships.)
Answer: 6
2014 Canadian Senior Mathematics Contest Solutions
Page 3
5. We use the following facts about exponents:
√
• 3 a = a1/3 for every real number a
• Every real number can be squared
• The cube root of every number exists
• (bc )d = bcd = bdc = (bd )c whenever bc and bd and bcd all exist
• (mn)p = mp np whenever mp and np and (mn)p all exist
Manipulating the given equation, we obtain
p
p
√
3
3
(2 + x)2 + 3 3 (2 − x)2 = 4 4 − x2
1/3
1/3
1/3
= 4 4 − x2
+ 3 (2 − x)2
(2 + x)2
(2 + x)2/3 + 3(2 − x)2/3 = 4 ((2 + x)(2 − x))1/3
2
2
= 4(2 + x)1/3 (2 − x)1/3
(2 + x)1/3 + 3 (2 − x)1/3
Letting u = (2 + x)1/3 and v = (2 − x)1/3 , this equation becomes
u2 + 3v 2 = 4uv
u2 − 4uv + 3v 2 = 0
(u − v)(u − 3v) = 0
Therefore, u = v or u = 3v.
If u = v, then (2 + x)1/3 = (2 − x)1/3 .
Cubing both sides gives 2 + x = 2 − x and so 2x = 0 or x = 0.
If u = 3v, then (2 + x)1/3 = 3(2 − x)1/3 .
Cubing both sides gives 2 + x = 33 (2 − x) and so 2 + x = 54 − 27x or 28x = 52 or x =
52
28
=
13
.
7
We verify these answers in the original equation.
If x = 0, the left side of the equation becomes
p
p
√
√
√
√
3
3
3
3
3
(2 + 0)2 + 3 3 (2 − 0)2 = 4 + 3 4 = 4 4 = 4 4 − 02
which equals the right side of the equation, and so x = 0 is a solution.
If x =
13
,
7
the right side of the equation equals
q
q
q
q
2
3
3 196−169
3 27
3
3
169
=
4 4 − 13
=
4
=
4
= 4√
=
4
4
−
3
7
49
49
49
49
and the left side of the equation equals
q
q
q
q
3
3
3
3
27 2
13 2
13 2
(2 + 7 ) + 3 (2 − 7 ) =
+3
7
1 2
7
=
q
3
729
49
q
+33
1
49
=
9
√
3
49
12
√
3
49
1
+3√
=
3
49
12
√
3
49
(We used the fact that 729 = 93 .) Therefore, the left side equals the right side, as required,
is also a solution.
and so x = 13
7
Thus, the solutions to the original equation are x = 0 and x =
13
.
7
Answer: 0, 13
7
2014 Canadian Senior Mathematics Contest Solutions
Page 4
6. If m and n are even and odd or odd and even, then m − n is odd and so the lockers numbered
m and n must have different colours.
Thus, none of the even-numbered lockers (we call these “even lockers”) can be painted the same
colour as any of the odd-numbered lockers (“odd lockers”).
(If m and n are both odd or both even, then m − n is even, so there is no restriction.)
This means that we cannot use all three colours in painting the even lockers and we cannot use
all three colours in painting the odd lockers, otherwise there would be no colours left to use for
the other set.
Thus, one or two colours are used to paint the even lockers and one or two colours are used to
paint the odd lockers.
Furthermore, we cannot use two colours to paint the even lockers and two colours to paint the
odd lockers, otherwise there would be a colour that overlaps.
Thus, either one colour is used for the even lockers and one colour for the odd lockers (it is
not required that all three colours be used), or one colour is used for the even lockers and two
colours for the odd lockers, or two colours are used for the even lockers and one colour for the
odd lockers.
We count the number of ways of painting in the first and second cases. The number of ways of
painting in the third case will be equal to the number of ways from the second case (the roles
of the even and odd lockers can be exchanged since there is an equal number of even lockers
and odd lockers).
Case 1: One colour for even lockers and one colour for odd lockers
There are three choices of colour for the even lockers (any of the three colours).
There are then two choices of colour for the odd lockers (either of the remaining colours).
In this case, there are 3 × 2 = 6 ways of colouring the lockers.
Case 2: One colour for even lockers and two colours for odd lockers
There are three choices of colour for the even lockers.
For each of these three choices, the odd lockers must be painted with the two remaining colours
(there is no choice of the colours), making sure to use each colour on at least one locker.
For each of the five odd lockers, there are two choices of colour, so there are 25 ways of painting
the odd lockers using at most these two colours. We subtract 2 ways from this total: one for
painting them all one colour and one for painting them all the other colour (since these was
counted in Case 1). Thus, there are 32 − 2 = 30 ways of painting the odd lockers using exactly
two colours.
In total, there are thus 3 × 30 = 90 ways of painting the lockers in this case.
Case 3: Two colours for even lockers and one colour for odd lockers
As in case 2, there are 90 ways of painting the lockers.
In total, there are 6 + 90 + 90 = 186 ways of painting the lockers.
Answer: 186
2014 Canadian Senior Mathematics Contest Solutions
Page 5
Part B
1. (a) From the given list, the numbers 5, 10, 15 form an arithmetic sequence with common
difference 5. None of the other sets of three numbers form an arithmetic sequence.
(b) Solution 1
Moving from the 2nd term to the 4th term involves adding the common difference twice.
Since the difference between the 2nd term and the 4th term is 13 − 7 = 6, then 2 times
the common difference equals 6, which means that the common difference is 3.
Therefore, the 3rd term, q, is 3 more than 7, or q = 10, and the 1st term, p, is 3 less than
7, or p = 4.
Thus, p = 4 and q = 10.
Solution 2
Suppose that the given arithmetic sequence has common difference d.
In terms of p and d, the four terms are p, p + d, p + 2d, p + 3d.
Since the four terms are p, 7, q, 13, then p + d = 7 and p + 3d = 13.
Subtracting the first of these equations from the second, we obtain 2d = 6 or d = 3.
Since p + d = 7 and d = 3, then p = 4.
Finally, q = p + 2d = 4 + 2(3) = 10.
Thus, p = 4 and q = 10.
(c) Solution 1
Moving from the 1st term to the 4th term involves adding the common difference 3 times.
Since the difference between the 1st term and the 4th term is (a + 21) − a = 21, then
3 times the common difference equals 21, which means that the common difference is 7.
The difference between the 3rd term, c, and the 1st term, a, is twice the common difference,
or 14, and so c − a = 14.
Solution 2
Suppose that the given arithmetic sequence has common difference d.
In terms of a and d, the four terms are a, a + d, a + 2d, a + 3d.
Since the four terms are a, b, c, a + 21, then a + 3d = a + 21 or 3d = 21, and so d = 7.
Since c = a + 2d, then c = a + 14.
Finally, c − a = (a + 14) − a = 14.
(d) Since (y−6), (2y+3), (y 2 +2) form an arithmetic sequence in that order, then the differences
between consecutive terms are equal and so
(2y + 3) − (y − 6)
y+9
0
0
=
=
=
=
(y 2 + 2) − (2y + 3)
y 2 − 2y − 1
y 2 − 3y − 10
(y − 5)(y + 2)
Therefore, y = 5 or y = −2.
We verify that each of these values of y gives an arithmetic sequence.
When y = 5, the sequence is −1, 13, 27 which is an arithmetic sequence with common
difference 14.
When y = −2, the sequence is −8, −1, 6 which is an arithmetic sequence with common
difference 7.
2014 Canadian Senior Mathematics Contest Solutions
Page 6
2. (a) (i) The semi-circles with diameter Y Z, XZ and XY have diameters of length x, y and
z, and so have radii of length 12 x, 12 y and 21 z, respectively.
Z
y
X
x
z
Y
In terms of x, the area of a semi-circle with radius 12 x is 12 π( 12 x)2 = 81 πx2 . (This is
one-half of the area of the circle with radius 12 x.)
Similarly, in terms of y and z, the areas of the semi-circles with diameter XZ and
XY are 18 πy 2 and 18 πz 2 , respectively.
Therefore, 18 πx2 = 50π and 81 πy 2 = 288π and we want to determine 18 πz 2 .
Since △XY Z is right-angled at Z, then Y Z 2 + XZ 2 = XY 2 , by the Pythagorean
Theorem, and so x2 + y 2 = z 2 .
Multiplying both sides of this equation by 18 π, we obtain 81 πx2 + 81 πy 2 = 18 πz 2 .
Therefore, 50π + 288π = 18 πz 2 and so 18 πz 2 = 338π.
Thus, the area of the semi-circle with diameter XY is 18 πz 2 = 338π.
(Note that we did not need to determine x, y or z.)
√
(ii) A square with side length s has diagonal of length 2s. This is because the diagonal
is the hypotenuse of an isosceles right-angled triangle with equal sides of length s.
U
b
T
R
q
r
P
a
p
Q
c
√
√
√
S
Therefore, a = 2p, b = 2q and c = 2r.
Since △P QR is right-angled at R, then QR2 + P R2 = P Q2 , by the Pythagorean
Theorem, and so p2 + q 2 = r2 .
Multiplying
sides
of this equation by 2, we obtain 2p2 + 2q 2 = 2r2 , and so
√ both
√
√ 2
( 2p) + ( 2q)2 = ( 2r)2 or a2 + b2 = c2 .
Thus, the lengths a, b and c satisfy the Pythagorean equation, which tells us that the
triangle formed with side lengths a, b and c is a right-angled triangle.
Note that the Pythagorean Theorem has two parts:
Suppose that △XY Z has side lengths Y Z = x, XZ = y and XY = z. Then:
• If the triangle is right-angled at Z, then x2 + y 2 = z 2 , and
• If x2 + y 2 = z 2 , then the triangle is right-angled at Z.
2014 Canadian Senior Mathematics Contest Solutions
Page 7
(b) Solution 1
Let AZ = h and let ∠ZXY = θ.
Since △XDE is right-angled at E, then
∠XDE = 180◦ − ∠DXE − ∠XED = 180◦ − θ − 90◦ = 90◦ − θ
Since ∠XDA = 90◦ , then ∠EDA = 90◦ − ∠XDE = 90◦ − (90◦ − θ) = θ.
Similarly, ∠EAD = 90◦ − θ, ∠DAZ = θ, ∠DZA = 90◦ − θ, ∠AZB = θ, ∠ZAB = 90◦ − θ,
and ∠BAC = θ.
Z
D
θ
θ
X
θ
θ
h
B
θ
E
A
C Y
Since AZ = h and △ADZ is right-angled at D, then AD = AZ cos(∠DAZ) = h cos θ.
Since AD = h cos θ and △DEA is right-angled at E, then
AE = AD sin(∠EDA) = (h cos θ) sin θ = h cos θ sin θ
Since AZ = h and △ZBA is right-angled at B, then AB = AZ sin(∠AZB) = h sin θ.
Since AB = h sin θ and △ACB is right-angled at C, then
AC = AB cos(∠BAC) = (h sin θ) cos θ = h cos θ sin θ
Therefore, AE = AC.
Solution 2
Let AZ = h and let ∠ZXY = θ.
Since △XDE is right-angled at E, then
∠XDE = 180◦ − ∠DXE − ∠XED = 180◦ − θ − 90◦ = 90◦ − θ
Since ∠XDA = 90◦ , then ∠EDA = 90◦ − ∠XDE = 90◦ − (90◦ − θ) = θ.
Similarly, ∠EAD = 90◦ − θ, ∠DAZ = θ, ∠DZA = 90◦ − θ, ∠AZB = θ, ∠ZAB = 90◦ − θ,
and ∠BAC = θ.
Construct rectangle JP CE with side JP passing through Z, J on ED extended and P
on CB extended.
Z
J
D
P
θ
θ
X
θ
θ
h
B
θ
E
A
C Y
Now ABZD, JZAE and ZP CA are all rectangles since they each have three right angles.
Therefore, JZ = EA. We will show that JZ = AC, which will tell us that AE = AC.
Since ABZD is a rectangle, then AB = DZ.
Since JZ is parallel to AC and DZ is parallel to AB, then ∠JZD = ∠BAC.
Since DJ is parallel to CB and DZ is parallel to AB, then ∠JDZ = ∠CBA.
Therefore, △ZDJ is congruent to △ABC (angle-side-angle).
Thus, JZ = CA and so AE = JZ = AC, as required.
2014 Canadian Senior Mathematics Contest Solutions
Page 8
√
3. (a) Suppose
that
x
=
a is a solution of x2 = 3⌊x⌋ + 1 for some positive √
integer a.
√
Since a ≥ 1,
√ then there exists a positive integer n for which n ≤ a < n + 1, which
means that ⌊ a⌋ = n.
√
Substituting into the equation x2 = 3⌊x⌋ +√
1, we obtain ( a)2 = 3n + 1 or a = 3n + 1.
Furthermore, squaring the inequality n ≤ a < n + 1, we obtain n2 ≤ a < (n + 1)2 or
n2 ≤ a < n2 + 2n + 1. (Since n > 0, the inequalities are preserved.)
Since a = 3n + 1 and n2 ≤ a < n2 + 2n + 1, then n2 ≤ 3n + 1 < n2 + 2n + 1, which gives
n2 − 3n − 1 ≤ 0 < n2 − n.
This inequality is equivalent to the statement that n2 − 3n − 1 ≤ 0 and 0 < n2 − n.
√
3 − 13
2
The roots of the quadratic equation n − 3n − 1 = 0 are n =
≈ −0.303 and
2
√
3 + 13
≈ 3.303.
n=
2
To have n2 − 3n − 1 ≤ 0, we want n to be between these two roots (since the function
f (n) = n2 − 3n − 1 represents a parabola opening upwards); since n is an integer, then
the interval narrows further to 0 ≤ n ≤ 3.
The roots of the quadratic equation n2 − n = 0 are n = 0 and n = 1.
To have n2 − n > 0, we want n to be “outside” these two roots; therefore, n < 0 or n > 1.
Thus, we need 0 ≤ n ≤ 3 and either n < 0 or n > 1.
Since n is an integer, then n = 2 or n = 3.
If n = 2, then a = 3n + 1 gives a = 7, which is the solution that we were given.
If n = 3, then a = 3n + 1 gives
√ a = 10, which is the solution that we want.
Checking, we see that if x = 10, then 3 ≤ x < 4, so 3⌊x⌋ + 1 = 3(3) + 1 = 10 = x2 , as
required.
Therefore, the value of a is 10.
(b) Suppose that x is a real number with ⌊x⌋ = n.
Then n ≤ x < n + 1.
Squaring each part, we obtain n2 ≤ x2 < n2 + 2n + 1; since n > 0, the inequalities are
preserved.
Therefore, n2 − 3⌊x⌋ ≤ x2 − 3⌊x⌋ < n2 + 2n + 1 − 3⌊x⌋.
Since ⌊x⌋ = n, then n2 −3n ≤ x2 −3⌊x⌋ < n2 +2n+1−3n or n2 −3n ≤ x2 −3⌊x⌋ < n2 −n+1.
Since x2 − 3⌊x⌋ is an integer, then n2 − 3n ≤ x2 − 3⌊x⌋ ≤ n2 − n.
Therefore, x2 − 3⌊x⌋ must lie between the integers n2 − 3n and n2 − n, inclusive.
2
We can
3⌊x⌋ can achieve each of these values by considering the values
√ show that 2x −
2
x = m for m = n , n + 1, . . . , n2 + 2n.
(c) First, we note that for any integer k with k ≥ 0, any
√ real number
√ x that is a solution to
the equation x2 = 3⌊x⌋ + k 2 − 1 is of the form x = a or x = − a for some non-negative
integer a. This is because if x is a solution, then the right side 3⌊x⌋ + k 2 − 1 is an integer,
√
so x2 is an integer which must be non-negative, say x2 = a, which means that x = ± a.
We break our solution into four cases: k = 0, k = 1, k = 2, and k ≥ 3.
Throughout this solution, a always represents a non-negative integer and n an integer.
Case 1: k = 0
We solve x2 = 3⌊x⌋ − 1.
√
First, we look for solutions
√ x = a with a ≥ 0.
√
Since x ≥ 0,
then
n
≤
a
<
n
+
1
for
some
non-negative
integer
n;
thus,
⌊x⌋
=
⌊
a⌋ = n.
√
√
Since x = a is a solution, then ( a)2 = 3n − 1, or a = 3n − 1.
2014 Canadian Senior Mathematics Contest Solutions
Page 9
√
Since n ≤ a < n + 1 and n > 0, then n2 ≤ a < n2 + 2n + 1.
2
Therefore, n2 ≤ 3n − 1 < n2 + 2n + 1 and √
so n2 − 3n + 1 ≤ 0 and n√
− n + 2 > 0.
3
+
5
5
3
−
≈ 0.382 and n =
≈ 2.618.
The roots of n2 − 3n + 1 = 0 are n =
2
2
√
√
3− 5
3+ 5
2
The solution to n − 3n + 1 ≤ 0 is
≤n≤
(n is between the roots).
2
2
Since n is an integer with n2 − 3n + 1 ≤ 0, then 1 ≤ n ≤ 2.
The equation n2 − n + 2 = 0 has no real roots and the corresponding parabola lies entirely
above the horizontal axis, so every integer n satisfies n2 − n + 2 > 0.
Therefore, the integers n that satisfy the inequalities n2 − 3n + 1 ≤ 0 and n2 − n + 2 > 0
are n = 1 and n = 2.
√
When n = 1, we have a = 3(1) − 1 = 2, and so x = √2 is a solution to the equation.
When n = 2, we have a = 3(2) − 1√= 5, and so x = 5 is a solution to the equation.
Next, we look for solutions
√ x = − a with a > 0.
Since x < 0, then
√ n ≤ − a < n + 1 for some negative integer n.
Thus, ⌊x⌋ =√
⌊− a⌋ = n.
√
Since x = − √a is a solution, then (− a)2 = 3n − 1, or a = 3n − 1.
Since n ≤ − a < n + 1 and n < 0, then n2 ≥ a > n2 + 2n + 1.
Therefore, n2 ≥ 3n − 1 > n2 + 2n + 1 and so n2 − 3n + 1 ≥ 0 and n2 − n + 2 < 0.
The equation n2 − n + 2 = 0 has no real roots and the corresponding parabola lies entirely
above the horizontal axis, so no integer n satisfies n2 − n + 2 < 0.
Therefore, there are no integers n that satisfy the√inequalities n2 − 3n + 1 ≥ 0 and
n2 − n + 2 < 0, and so there are no solutions √
x = − a. √
Therefore, when k = 0, the solutions are x = 2 and x = 5.
Case 2: k = 1
We solve x2 = 3⌊x⌋.
√
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n and the inequalities n2 −3n ≤ 0 and n2 −n+1 > 0.
The first inequality has integer solutions n = 0, 1, 2, 3 and the second inequality is true for
all integers n.
√ √
√ √ √ √
or
x
=
0,
3, 6, 3.
Using n = 0, 1, 2, 3, we obtain the √
solutions x = 0, 3, 6, 9,
√
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n.
Proceeding as above, we obtain a = 3n and the inequalities n2 −3n ≥ 0 and n2 −n+1 < 0.
The inequality n2 − n + 1 < 0 is not true for any integer n, so there are no solutions in
this case.
√ √
Therefore, when k = 1, the solutions are x = 0, 3, 6, 3.
Case 3: k = 2
We solve x2 = 3⌊x⌋ + 3.
√
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n + 3 and the inequalities n2 − 3n − 3 ≤ 0 and
n2 − n − 2 > 0.
The first inequality has integer solutions n = 0, 1, 2, 3 and the second inequality is true
when n < −1 or n > 2.
Therefore, these two inequalities are both true only when
√ n = 3.
When n = 3, we obtain a = 3(3) +√3 = 12, and so x = 12 is a√solution to the equation.
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n.
Proceeding as above, we obtain a = 3n + 3 and the inequalities n2 − 3n − 3 ≥ 0 and
n2 − n − 2 < 0.
2014 Canadian Senior Mathematics Contest Solutions
Page 10
The second inequality is not true for any negative integer n (since its real number solution
is −1 < n < 2), so there are no solutions in√this case.
Therefore, when k = 2, the solution is x = 12.
Case 4: k ≥ 3
We solve x2 = 3⌊x⌋ + (k 2 − 1). √
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n+(k 2 −1) and the inequalities n2 −3n−(k 2 −1) ≤ 0
and n2 − n − (k 2 − 2) > 0.
Consider the functions
f (n) = n2 − 3n − (k 2 − 1)
g(n) = n2 − n − (k 2 − 2)
Note that k 2 − 1 > 0 and k 2 − 2 > 0 since k ≥ 3.
Since the leading coefficient is positive and the constant term is negative in each case,
then each of f (n) and g(n) has a positive zero and a negative zero.
Therefore, the integer solution set to the inequality n2 − 3n − (k 2 − 1) ≤ 0 will be of the
form C ≤ n ≤ D for some integers C ≤ 0 ≤ D. (Note that the real number solution set
to the inequality n2 − 3n − (k 2 − 1) ≤ 0 is of the form c ≤ n ≤ d for some real numbers
c < 0 < d. When we restrict this set to integers, the solution set is of the form C ≤ n ≤ D
with C ≤ 0 ≤ D, where C is the smallest integer larger than c and D is the largest integer
less than d.)
Also, the integer solution set to the inequality n2 − n − (k 2 − 2) > 0 will be of the form
n ≤ E or n ≥ F for some integers E < 0 < F . (Note that the real number solution set to
the inequality n2 − n − (k 2 − 2) > 0 is of the form n < e or n > f for some real numbers
e < 0 < f . When we restrict this set to integers, the solution set is of the form n ≤ E
or n ≥ F , where E is the largest integer less than e and F is the smallest integer greater
than f .)
Since we are currently restricting to the case that n ≥ 0, then the set of integers n that
satisfy these inequalities is of the form 0 ≤ n ≤ D and n ≥ F .
We show that n = k + 1 satisfies both inequalities, that n = k + 2 does not satisfy the first
inequality (thus D = k + 1), and that n = k does not satisfy the second inequality (thus
F = k + 1). Combining this information will tell us that n = k + 1 is the unique positive
integer that satisfies these restrictions.
Now
f (k + 1)
f (k + 2)
g(k + 1)
g(k)
=
=
=
=
(k + 1)2 − 3(k + 1) − (k 2 − 1) = k 2 + 2k + 1 − 3k − 3 − k 2 + 1 = −k − 1 < 0
(k + 2)2 − 3(k + 2) − (k 2 − 1) = k 2 + 4k + 4 − 3k − 6 − k 2 + 1 = k − 1 > 0
(k + 1)2 − (k + 1) − (k 2 − 2) = k 2 + 2k + 1 − k − 1 − k 2 + 2 = k + 2 > 0
k 2 − k − (k 2 − 2) = −k + 2 < 0
(The final inequalities in each case come from the fact that k ≥ 3.)
Therefore, n = k + 1 is the unique non-negative integer that satisfies the inequalities.
Since√n = √
k + 1, then a = 3n + (k 2 − 1) = 3(k + 1) + (k 2 − 1) = k 2 + 3k + 2 and so
x = a = k 2 + 3k + 2.
√
√
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n with n < 0.
Proceeding as above, we obtain a = 3n+(k 2 −1) and the inequalities n2 −3n−(k 2 −1) ≥ 0
and n2 − n − (k 2 − 2) < 0.
Modelling the section of this case when n ≥ 0, we find that the solution to the first
inequality is of the form n ≤ G or n ≥ H for some integers G < 0 < H and the solution
2014 Canadian Senior Mathematics Contest Solutions
Page 11
to the second inequality is of the form J ≤ n ≤ L for some integers J ≤ 0 ≤ L.
Since we are considering the case that n < 0, then the set of integers n that satisfy these
inequalities is of the form n ≤ G and J ≤ n < 0.
We show that n = −k + 1 satisfies both inequalities, that n = −k + 2 does not satisfy the
first inequality (thus G = −k + 1), and that n = −k does not satisfy the second inequality
(thus J = −k + 1). Combining this information will tell us that n = −k + 1 is the unique
possible negative value.
Now
f (−k + 1)
f (−k + 2)
g(−k + 1)
g(k)
=
=
=
=
(−k + 1)2 − 3(−k + 1) − (k 2 − 1) = k − 1 > 0
(−k + 2)2 − 3(−k + 2) − (k 2 − 1) = −k − 1 < 0
(−k + 1)2 − (−k + 1) − (k 2 − 2) = −k + 2 < 0
(−k)2 − (−k) − (k 2 − 2) = k + 2 > 0
(The final inequalities in each case come from the fact that k ≥ 3.)
Therefore, n = −k + 1 is the unique negative integer that satisfies the inequalities.
Since n
+ 1, then a = 3n + (k 2 − 1) = 3(−k + 1) + (k 2 − 1) = k 2 − 3k + 2 and so
√= −k √
x = − a = − k 2 − 3k + 2.
√
√
Therefore, in the case that k ≥ 3, the solutions are x = k 2 + 3k + 2 and x = − k 2 − 3k + 2.
In summary, the solutions are given in the following table:
k=0
k=1
k=2
k≥3
x
√ √
2, 5
√ √
0, 3, 6, 3
√
12
√
√
k 2 + 3k + 2, − k 2 − 3k + 2
in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
2014 Canadian Senior
Mathematics Contest
Thursday, November 20, 2014
(in North America and South America)
Friday, November 21, 2014
(outside of North America and South America)
Solutions
©2014 University of Waterloo
2014 Canadian Senior Mathematics Contest Solutions
Page 2
Part A
1. We note that ∠DAF = ∠F AB − ∠DAB and that ∠F AB = ∠F AE + ∠EAB.
Since ABCD is a square, then ∠DAB = 90◦ .
Since each of △AEF and △ABE is equilateral, then ∠F AE = ∠EAB = 60◦ .
Therefore, ∠DAF = (60◦ + 60◦ ) − 90◦ = 30◦ .
Answer: 30◦
2. Since the ratio of the number of dimes to the number of quarters is 3 : 2, then we let the
numbers of dimes and quarters be 3k and 2k, respectively, for some positive integer k.
Since each dime is worth 10 cents, then the 3k dimes are worth 10(3k) = 30k cents.
Since each quarter is worth 25 cents, then the 2k quarters are worth 25(2k) = 50k cents.
Since the total value of the coins is $4 which equals 400 cents, then 30k + 50k = 400 or
80k = 400, and so k = 5.
Therefore, there are 3(5) = 15 dimes in the jar.
(Note that 15 dimes are worth 150 cents and 2(5) = 10 quarters are worth 250 cents, for a total
of 400 cents, as required.)
Answer: 15
3. We note that 5000 = 5 · 1000 = 5 · 103 = 5 · (2 · 5)3 = 5 · 23 · 53 = 23 54 .
Since neither m nor n is divisible by 10, then neither m nor n can have factors of both 2 and 5.
Since the product mn equals 5000, the only prime factors of 5000 are 2 and 5, and neither m
nor n has factors of both 2 and 5, then m and n must be 23 and 54 , or 54 and 23 .
Therefore, m + n = 23 + 54 = 8 + 625 = 633.
Answer: 633
4. Solution 1
Since f (x) + f (x + 3) = 2x + 5 for all x, then using x = 2 we obtain f (2) + f (5) = 2(2) + 5
and so f (2) + f (5) = 9.
Also, using x = 5 we obtain f (5) + f (8) = 2(5) + 5, and so f (5) + f (8) = 15.
Adding these two equations, we obtain f (2) + f (8) + 2f (5) = 24.
Since f (8) + f (2) = 12, then 2f (5) = 24 − (f (2) + f (8)) = 24 − 12 = 12 and so f (5) = 6.
Solution 2
The function f (x) = x + 1 satisfies the required properties, since
• f (x + 3) = (x + 3) + 1 = x + 4 and so f (x) + f (x + 3) = (x + 1) + (x + 4) = 2x + 5, and
• f (2) = 2 + 1 = 3 and f (8) = 8 + 1 = 9 gives f (8) + f (2) = 12.
So f (x) = x + 1 is a function that works and the question implies that the value of f (5) is the
same no matter which function f that satisfies the given properties is chosen.
Therefore, f (5) = 5 + 1 = 6.
(Note that there may be other functions that satisfy these relationships.)
Answer: 6
2014 Canadian Senior Mathematics Contest Solutions
Page 3
5. We use the following facts about exponents:
√
• 3 a = a1/3 for every real number a
• Every real number can be squared
• The cube root of every number exists
• (bc )d = bcd = bdc = (bd )c whenever bc and bd and bcd all exist
• (mn)p = mp np whenever mp and np and (mn)p all exist
Manipulating the given equation, we obtain
p
p
√
3
3
(2 + x)2 + 3 3 (2 − x)2 = 4 4 − x2
1/3
1/3
1/3
= 4 4 − x2
+ 3 (2 − x)2
(2 + x)2
(2 + x)2/3 + 3(2 − x)2/3 = 4 ((2 + x)(2 − x))1/3
2
2
= 4(2 + x)1/3 (2 − x)1/3
(2 + x)1/3 + 3 (2 − x)1/3
Letting u = (2 + x)1/3 and v = (2 − x)1/3 , this equation becomes
u2 + 3v 2 = 4uv
u2 − 4uv + 3v 2 = 0
(u − v)(u − 3v) = 0
Therefore, u = v or u = 3v.
If u = v, then (2 + x)1/3 = (2 − x)1/3 .
Cubing both sides gives 2 + x = 2 − x and so 2x = 0 or x = 0.
If u = 3v, then (2 + x)1/3 = 3(2 − x)1/3 .
Cubing both sides gives 2 + x = 33 (2 − x) and so 2 + x = 54 − 27x or 28x = 52 or x =
52
28
=
13
.
7
We verify these answers in the original equation.
If x = 0, the left side of the equation becomes
p
p
√
√
√
√
3
3
3
3
3
(2 + 0)2 + 3 3 (2 − 0)2 = 4 + 3 4 = 4 4 = 4 4 − 02
which equals the right side of the equation, and so x = 0 is a solution.
If x =
13
,
7
the right side of the equation equals
q
q
q
q
2
3
3 196−169
3 27
3
3
169
=
4 4 − 13
=
4
=
4
= 4√
=
4
4
−
3
7
49
49
49
49
and the left side of the equation equals
q
q
q
q
3
3
3
3
27 2
13 2
13 2
(2 + 7 ) + 3 (2 − 7 ) =
+3
7
1 2
7
=
q
3
729
49
q
+33
1
49
=
9
√
3
49
12
√
3
49
1
+3√
=
3
49
12
√
3
49
(We used the fact that 729 = 93 .) Therefore, the left side equals the right side, as required,
is also a solution.
and so x = 13
7
Thus, the solutions to the original equation are x = 0 and x =
13
.
7
Answer: 0, 13
7
2014 Canadian Senior Mathematics Contest Solutions
Page 4
6. If m and n are even and odd or odd and even, then m − n is odd and so the lockers numbered
m and n must have different colours.
Thus, none of the even-numbered lockers (we call these “even lockers”) can be painted the same
colour as any of the odd-numbered lockers (“odd lockers”).
(If m and n are both odd or both even, then m − n is even, so there is no restriction.)
This means that we cannot use all three colours in painting the even lockers and we cannot use
all three colours in painting the odd lockers, otherwise there would be no colours left to use for
the other set.
Thus, one or two colours are used to paint the even lockers and one or two colours are used to
paint the odd lockers.
Furthermore, we cannot use two colours to paint the even lockers and two colours to paint the
odd lockers, otherwise there would be a colour that overlaps.
Thus, either one colour is used for the even lockers and one colour for the odd lockers (it is
not required that all three colours be used), or one colour is used for the even lockers and two
colours for the odd lockers, or two colours are used for the even lockers and one colour for the
odd lockers.
We count the number of ways of painting in the first and second cases. The number of ways of
painting in the third case will be equal to the number of ways from the second case (the roles
of the even and odd lockers can be exchanged since there is an equal number of even lockers
and odd lockers).
Case 1: One colour for even lockers and one colour for odd lockers
There are three choices of colour for the even lockers (any of the three colours).
There are then two choices of colour for the odd lockers (either of the remaining colours).
In this case, there are 3 × 2 = 6 ways of colouring the lockers.
Case 2: One colour for even lockers and two colours for odd lockers
There are three choices of colour for the even lockers.
For each of these three choices, the odd lockers must be painted with the two remaining colours
(there is no choice of the colours), making sure to use each colour on at least one locker.
For each of the five odd lockers, there are two choices of colour, so there are 25 ways of painting
the odd lockers using at most these two colours. We subtract 2 ways from this total: one for
painting them all one colour and one for painting them all the other colour (since these was
counted in Case 1). Thus, there are 32 − 2 = 30 ways of painting the odd lockers using exactly
two colours.
In total, there are thus 3 × 30 = 90 ways of painting the lockers in this case.
Case 3: Two colours for even lockers and one colour for odd lockers
As in case 2, there are 90 ways of painting the lockers.
In total, there are 6 + 90 + 90 = 186 ways of painting the lockers.
Answer: 186
2014 Canadian Senior Mathematics Contest Solutions
Page 5
Part B
1. (a) From the given list, the numbers 5, 10, 15 form an arithmetic sequence with common
difference 5. None of the other sets of three numbers form an arithmetic sequence.
(b) Solution 1
Moving from the 2nd term to the 4th term involves adding the common difference twice.
Since the difference between the 2nd term and the 4th term is 13 − 7 = 6, then 2 times
the common difference equals 6, which means that the common difference is 3.
Therefore, the 3rd term, q, is 3 more than 7, or q = 10, and the 1st term, p, is 3 less than
7, or p = 4.
Thus, p = 4 and q = 10.
Solution 2
Suppose that the given arithmetic sequence has common difference d.
In terms of p and d, the four terms are p, p + d, p + 2d, p + 3d.
Since the four terms are p, 7, q, 13, then p + d = 7 and p + 3d = 13.
Subtracting the first of these equations from the second, we obtain 2d = 6 or d = 3.
Since p + d = 7 and d = 3, then p = 4.
Finally, q = p + 2d = 4 + 2(3) = 10.
Thus, p = 4 and q = 10.
(c) Solution 1
Moving from the 1st term to the 4th term involves adding the common difference 3 times.
Since the difference between the 1st term and the 4th term is (a + 21) − a = 21, then
3 times the common difference equals 21, which means that the common difference is 7.
The difference between the 3rd term, c, and the 1st term, a, is twice the common difference,
or 14, and so c − a = 14.
Solution 2
Suppose that the given arithmetic sequence has common difference d.
In terms of a and d, the four terms are a, a + d, a + 2d, a + 3d.
Since the four terms are a, b, c, a + 21, then a + 3d = a + 21 or 3d = 21, and so d = 7.
Since c = a + 2d, then c = a + 14.
Finally, c − a = (a + 14) − a = 14.
(d) Since (y−6), (2y+3), (y 2 +2) form an arithmetic sequence in that order, then the differences
between consecutive terms are equal and so
(2y + 3) − (y − 6)
y+9
0
0
=
=
=
=
(y 2 + 2) − (2y + 3)
y 2 − 2y − 1
y 2 − 3y − 10
(y − 5)(y + 2)
Therefore, y = 5 or y = −2.
We verify that each of these values of y gives an arithmetic sequence.
When y = 5, the sequence is −1, 13, 27 which is an arithmetic sequence with common
difference 14.
When y = −2, the sequence is −8, −1, 6 which is an arithmetic sequence with common
difference 7.
2014 Canadian Senior Mathematics Contest Solutions
Page 6
2. (a) (i) The semi-circles with diameter Y Z, XZ and XY have diameters of length x, y and
z, and so have radii of length 12 x, 12 y and 21 z, respectively.
Z
y
X
x
z
Y
In terms of x, the area of a semi-circle with radius 12 x is 12 π( 12 x)2 = 81 πx2 . (This is
one-half of the area of the circle with radius 12 x.)
Similarly, in terms of y and z, the areas of the semi-circles with diameter XZ and
XY are 18 πy 2 and 18 πz 2 , respectively.
Therefore, 18 πx2 = 50π and 81 πy 2 = 288π and we want to determine 18 πz 2 .
Since △XY Z is right-angled at Z, then Y Z 2 + XZ 2 = XY 2 , by the Pythagorean
Theorem, and so x2 + y 2 = z 2 .
Multiplying both sides of this equation by 18 π, we obtain 81 πx2 + 81 πy 2 = 18 πz 2 .
Therefore, 50π + 288π = 18 πz 2 and so 18 πz 2 = 338π.
Thus, the area of the semi-circle with diameter XY is 18 πz 2 = 338π.
(Note that we did not need to determine x, y or z.)
√
(ii) A square with side length s has diagonal of length 2s. This is because the diagonal
is the hypotenuse of an isosceles right-angled triangle with equal sides of length s.
U
b
T
R
q
r
P
a
p
Q
c
√
√
√
S
Therefore, a = 2p, b = 2q and c = 2r.
Since △P QR is right-angled at R, then QR2 + P R2 = P Q2 , by the Pythagorean
Theorem, and so p2 + q 2 = r2 .
Multiplying
sides
of this equation by 2, we obtain 2p2 + 2q 2 = 2r2 , and so
√ both
√
√ 2
( 2p) + ( 2q)2 = ( 2r)2 or a2 + b2 = c2 .
Thus, the lengths a, b and c satisfy the Pythagorean equation, which tells us that the
triangle formed with side lengths a, b and c is a right-angled triangle.
Note that the Pythagorean Theorem has two parts:
Suppose that △XY Z has side lengths Y Z = x, XZ = y and XY = z. Then:
• If the triangle is right-angled at Z, then x2 + y 2 = z 2 , and
• If x2 + y 2 = z 2 , then the triangle is right-angled at Z.
2014 Canadian Senior Mathematics Contest Solutions
Page 7
(b) Solution 1
Let AZ = h and let ∠ZXY = θ.
Since △XDE is right-angled at E, then
∠XDE = 180◦ − ∠DXE − ∠XED = 180◦ − θ − 90◦ = 90◦ − θ
Since ∠XDA = 90◦ , then ∠EDA = 90◦ − ∠XDE = 90◦ − (90◦ − θ) = θ.
Similarly, ∠EAD = 90◦ − θ, ∠DAZ = θ, ∠DZA = 90◦ − θ, ∠AZB = θ, ∠ZAB = 90◦ − θ,
and ∠BAC = θ.
Z
D
θ
θ
X
θ
θ
h
B
θ
E
A
C Y
Since AZ = h and △ADZ is right-angled at D, then AD = AZ cos(∠DAZ) = h cos θ.
Since AD = h cos θ and △DEA is right-angled at E, then
AE = AD sin(∠EDA) = (h cos θ) sin θ = h cos θ sin θ
Since AZ = h and △ZBA is right-angled at B, then AB = AZ sin(∠AZB) = h sin θ.
Since AB = h sin θ and △ACB is right-angled at C, then
AC = AB cos(∠BAC) = (h sin θ) cos θ = h cos θ sin θ
Therefore, AE = AC.
Solution 2
Let AZ = h and let ∠ZXY = θ.
Since △XDE is right-angled at E, then
∠XDE = 180◦ − ∠DXE − ∠XED = 180◦ − θ − 90◦ = 90◦ − θ
Since ∠XDA = 90◦ , then ∠EDA = 90◦ − ∠XDE = 90◦ − (90◦ − θ) = θ.
Similarly, ∠EAD = 90◦ − θ, ∠DAZ = θ, ∠DZA = 90◦ − θ, ∠AZB = θ, ∠ZAB = 90◦ − θ,
and ∠BAC = θ.
Construct rectangle JP CE with side JP passing through Z, J on ED extended and P
on CB extended.
Z
J
D
P
θ
θ
X
θ
θ
h
B
θ
E
A
C Y
Now ABZD, JZAE and ZP CA are all rectangles since they each have three right angles.
Therefore, JZ = EA. We will show that JZ = AC, which will tell us that AE = AC.
Since ABZD is a rectangle, then AB = DZ.
Since JZ is parallel to AC and DZ is parallel to AB, then ∠JZD = ∠BAC.
Since DJ is parallel to CB and DZ is parallel to AB, then ∠JDZ = ∠CBA.
Therefore, △ZDJ is congruent to △ABC (angle-side-angle).
Thus, JZ = CA and so AE = JZ = AC, as required.
2014 Canadian Senior Mathematics Contest Solutions
Page 8
√
3. (a) Suppose
that
x
=
a is a solution of x2 = 3⌊x⌋ + 1 for some positive √
integer a.
√
Since a ≥ 1,
√ then there exists a positive integer n for which n ≤ a < n + 1, which
means that ⌊ a⌋ = n.
√
Substituting into the equation x2 = 3⌊x⌋ +√
1, we obtain ( a)2 = 3n + 1 or a = 3n + 1.
Furthermore, squaring the inequality n ≤ a < n + 1, we obtain n2 ≤ a < (n + 1)2 or
n2 ≤ a < n2 + 2n + 1. (Since n > 0, the inequalities are preserved.)
Since a = 3n + 1 and n2 ≤ a < n2 + 2n + 1, then n2 ≤ 3n + 1 < n2 + 2n + 1, which gives
n2 − 3n − 1 ≤ 0 < n2 − n.
This inequality is equivalent to the statement that n2 − 3n − 1 ≤ 0 and 0 < n2 − n.
√
3 − 13
2
The roots of the quadratic equation n − 3n − 1 = 0 are n =
≈ −0.303 and
2
√
3 + 13
≈ 3.303.
n=
2
To have n2 − 3n − 1 ≤ 0, we want n to be between these two roots (since the function
f (n) = n2 − 3n − 1 represents a parabola opening upwards); since n is an integer, then
the interval narrows further to 0 ≤ n ≤ 3.
The roots of the quadratic equation n2 − n = 0 are n = 0 and n = 1.
To have n2 − n > 0, we want n to be “outside” these two roots; therefore, n < 0 or n > 1.
Thus, we need 0 ≤ n ≤ 3 and either n < 0 or n > 1.
Since n is an integer, then n = 2 or n = 3.
If n = 2, then a = 3n + 1 gives a = 7, which is the solution that we were given.
If n = 3, then a = 3n + 1 gives
√ a = 10, which is the solution that we want.
Checking, we see that if x = 10, then 3 ≤ x < 4, so 3⌊x⌋ + 1 = 3(3) + 1 = 10 = x2 , as
required.
Therefore, the value of a is 10.
(b) Suppose that x is a real number with ⌊x⌋ = n.
Then n ≤ x < n + 1.
Squaring each part, we obtain n2 ≤ x2 < n2 + 2n + 1; since n > 0, the inequalities are
preserved.
Therefore, n2 − 3⌊x⌋ ≤ x2 − 3⌊x⌋ < n2 + 2n + 1 − 3⌊x⌋.
Since ⌊x⌋ = n, then n2 −3n ≤ x2 −3⌊x⌋ < n2 +2n+1−3n or n2 −3n ≤ x2 −3⌊x⌋ < n2 −n+1.
Since x2 − 3⌊x⌋ is an integer, then n2 − 3n ≤ x2 − 3⌊x⌋ ≤ n2 − n.
Therefore, x2 − 3⌊x⌋ must lie between the integers n2 − 3n and n2 − n, inclusive.
2
We can
3⌊x⌋ can achieve each of these values by considering the values
√ show that 2x −
2
x = m for m = n , n + 1, . . . , n2 + 2n.
(c) First, we note that for any integer k with k ≥ 0, any
√ real number
√ x that is a solution to
the equation x2 = 3⌊x⌋ + k 2 − 1 is of the form x = a or x = − a for some non-negative
integer a. This is because if x is a solution, then the right side 3⌊x⌋ + k 2 − 1 is an integer,
√
so x2 is an integer which must be non-negative, say x2 = a, which means that x = ± a.
We break our solution into four cases: k = 0, k = 1, k = 2, and k ≥ 3.
Throughout this solution, a always represents a non-negative integer and n an integer.
Case 1: k = 0
We solve x2 = 3⌊x⌋ − 1.
√
First, we look for solutions
√ x = a with a ≥ 0.
√
Since x ≥ 0,
then
n
≤
a
<
n
+
1
for
some
non-negative
integer
n;
thus,
⌊x⌋
=
⌊
a⌋ = n.
√
√
Since x = a is a solution, then ( a)2 = 3n − 1, or a = 3n − 1.
2014 Canadian Senior Mathematics Contest Solutions
Page 9
√
Since n ≤ a < n + 1 and n > 0, then n2 ≤ a < n2 + 2n + 1.
2
Therefore, n2 ≤ 3n − 1 < n2 + 2n + 1 and √
so n2 − 3n + 1 ≤ 0 and n√
− n + 2 > 0.
3
+
5
5
3
−
≈ 0.382 and n =
≈ 2.618.
The roots of n2 − 3n + 1 = 0 are n =
2
2
√
√
3− 5
3+ 5
2
The solution to n − 3n + 1 ≤ 0 is
≤n≤
(n is between the roots).
2
2
Since n is an integer with n2 − 3n + 1 ≤ 0, then 1 ≤ n ≤ 2.
The equation n2 − n + 2 = 0 has no real roots and the corresponding parabola lies entirely
above the horizontal axis, so every integer n satisfies n2 − n + 2 > 0.
Therefore, the integers n that satisfy the inequalities n2 − 3n + 1 ≤ 0 and n2 − n + 2 > 0
are n = 1 and n = 2.
√
When n = 1, we have a = 3(1) − 1 = 2, and so x = √2 is a solution to the equation.
When n = 2, we have a = 3(2) − 1√= 5, and so x = 5 is a solution to the equation.
Next, we look for solutions
√ x = − a with a > 0.
Since x < 0, then
√ n ≤ − a < n + 1 for some negative integer n.
Thus, ⌊x⌋ =√
⌊− a⌋ = n.
√
Since x = − √a is a solution, then (− a)2 = 3n − 1, or a = 3n − 1.
Since n ≤ − a < n + 1 and n < 0, then n2 ≥ a > n2 + 2n + 1.
Therefore, n2 ≥ 3n − 1 > n2 + 2n + 1 and so n2 − 3n + 1 ≥ 0 and n2 − n + 2 < 0.
The equation n2 − n + 2 = 0 has no real roots and the corresponding parabola lies entirely
above the horizontal axis, so no integer n satisfies n2 − n + 2 < 0.
Therefore, there are no integers n that satisfy the√inequalities n2 − 3n + 1 ≥ 0 and
n2 − n + 2 < 0, and so there are no solutions √
x = − a. √
Therefore, when k = 0, the solutions are x = 2 and x = 5.
Case 2: k = 1
We solve x2 = 3⌊x⌋.
√
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n and the inequalities n2 −3n ≤ 0 and n2 −n+1 > 0.
The first inequality has integer solutions n = 0, 1, 2, 3 and the second inequality is true for
all integers n.
√ √
√ √ √ √
or
x
=
0,
3, 6, 3.
Using n = 0, 1, 2, 3, we obtain the √
solutions x = 0, 3, 6, 9,
√
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n.
Proceeding as above, we obtain a = 3n and the inequalities n2 −3n ≥ 0 and n2 −n+1 < 0.
The inequality n2 − n + 1 < 0 is not true for any integer n, so there are no solutions in
this case.
√ √
Therefore, when k = 1, the solutions are x = 0, 3, 6, 3.
Case 3: k = 2
We solve x2 = 3⌊x⌋ + 3.
√
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n + 3 and the inequalities n2 − 3n − 3 ≤ 0 and
n2 − n − 2 > 0.
The first inequality has integer solutions n = 0, 1, 2, 3 and the second inequality is true
when n < −1 or n > 2.
Therefore, these two inequalities are both true only when
√ n = 3.
When n = 3, we obtain a = 3(3) +√3 = 12, and so x = 12 is a√solution to the equation.
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n.
Proceeding as above, we obtain a = 3n + 3 and the inequalities n2 − 3n − 3 ≥ 0 and
n2 − n − 2 < 0.
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The second inequality is not true for any negative integer n (since its real number solution
is −1 < n < 2), so there are no solutions in√this case.
Therefore, when k = 2, the solution is x = 12.
Case 4: k ≥ 3
We solve x2 = 3⌊x⌋ + (k 2 − 1). √
√
First, we look for solutions x = a with ⌊x⌋ = ⌊ a⌋ = n for some non-negative integer n.
Proceeding as above, we obtain a = 3n+(k 2 −1) and the inequalities n2 −3n−(k 2 −1) ≤ 0
and n2 − n − (k 2 − 2) > 0.
Consider the functions
f (n) = n2 − 3n − (k 2 − 1)
g(n) = n2 − n − (k 2 − 2)
Note that k 2 − 1 > 0 and k 2 − 2 > 0 since k ≥ 3.
Since the leading coefficient is positive and the constant term is negative in each case,
then each of f (n) and g(n) has a positive zero and a negative zero.
Therefore, the integer solution set to the inequality n2 − 3n − (k 2 − 1) ≤ 0 will be of the
form C ≤ n ≤ D for some integers C ≤ 0 ≤ D. (Note that the real number solution set
to the inequality n2 − 3n − (k 2 − 1) ≤ 0 is of the form c ≤ n ≤ d for some real numbers
c < 0 < d. When we restrict this set to integers, the solution set is of the form C ≤ n ≤ D
with C ≤ 0 ≤ D, where C is the smallest integer larger than c and D is the largest integer
less than d.)
Also, the integer solution set to the inequality n2 − n − (k 2 − 2) > 0 will be of the form
n ≤ E or n ≥ F for some integers E < 0 < F . (Note that the real number solution set to
the inequality n2 − n − (k 2 − 2) > 0 is of the form n < e or n > f for some real numbers
e < 0 < f . When we restrict this set to integers, the solution set is of the form n ≤ E
or n ≥ F , where E is the largest integer less than e and F is the smallest integer greater
than f .)
Since we are currently restricting to the case that n ≥ 0, then the set of integers n that
satisfy these inequalities is of the form 0 ≤ n ≤ D and n ≥ F .
We show that n = k + 1 satisfies both inequalities, that n = k + 2 does not satisfy the first
inequality (thus D = k + 1), and that n = k does not satisfy the second inequality (thus
F = k + 1). Combining this information will tell us that n = k + 1 is the unique positive
integer that satisfies these restrictions.
Now
f (k + 1)
f (k + 2)
g(k + 1)
g(k)
=
=
=
=
(k + 1)2 − 3(k + 1) − (k 2 − 1) = k 2 + 2k + 1 − 3k − 3 − k 2 + 1 = −k − 1 < 0
(k + 2)2 − 3(k + 2) − (k 2 − 1) = k 2 + 4k + 4 − 3k − 6 − k 2 + 1 = k − 1 > 0
(k + 1)2 − (k + 1) − (k 2 − 2) = k 2 + 2k + 1 − k − 1 − k 2 + 2 = k + 2 > 0
k 2 − k − (k 2 − 2) = −k + 2 < 0
(The final inequalities in each case come from the fact that k ≥ 3.)
Therefore, n = k + 1 is the unique non-negative integer that satisfies the inequalities.
Since√n = √
k + 1, then a = 3n + (k 2 − 1) = 3(k + 1) + (k 2 − 1) = k 2 + 3k + 2 and so
x = a = k 2 + 3k + 2.
√
√
Next, we look for solutions x = − a with a > 0 and ⌊x⌋ = ⌊− a⌋ = n with n < 0.
Proceeding as above, we obtain a = 3n+(k 2 −1) and the inequalities n2 −3n−(k 2 −1) ≥ 0
and n2 − n − (k 2 − 2) < 0.
Modelling the section of this case when n ≥ 0, we find that the solution to the first
inequality is of the form n ≤ G or n ≥ H for some integers G < 0 < H and the solution
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to the second inequality is of the form J ≤ n ≤ L for some integers J ≤ 0 ≤ L.
Since we are considering the case that n < 0, then the set of integers n that satisfy these
inequalities is of the form n ≤ G and J ≤ n < 0.
We show that n = −k + 1 satisfies both inequalities, that n = −k + 2 does not satisfy the
first inequality (thus G = −k + 1), and that n = −k does not satisfy the second inequality
(thus J = −k + 1). Combining this information will tell us that n = −k + 1 is the unique
possible negative value.
Now
f (−k + 1)
f (−k + 2)
g(−k + 1)
g(k)
=
=
=
=
(−k + 1)2 − 3(−k + 1) − (k 2 − 1) = k − 1 > 0
(−k + 2)2 − 3(−k + 2) − (k 2 − 1) = −k − 1 < 0
(−k + 1)2 − (−k + 1) − (k 2 − 2) = −k + 2 < 0
(−k)2 − (−k) − (k 2 − 2) = k + 2 > 0
(The final inequalities in each case come from the fact that k ≥ 3.)
Therefore, n = −k + 1 is the unique negative integer that satisfies the inequalities.
Since n
+ 1, then a = 3n + (k 2 − 1) = 3(−k + 1) + (k 2 − 1) = k 2 − 3k + 2 and so
√= −k √
x = − a = − k 2 − 3k + 2.
√
√
Therefore, in the case that k ≥ 3, the solutions are x = k 2 + 3k + 2 and x = − k 2 − 3k + 2.
In summary, the solutions are given in the following table:
k=0
k=1
k=2
k≥3
x
√ √
2, 5
√ √
0, 3, 6, 3
√
12
√
√
k 2 + 3k + 2, − k 2 − 3k + 2